You have 3.00 L of a 3.00 M solution of NaCl(aq) called solution A. You also have 2.00 L of a 2.00 M solution of AgNO3(aq) called solution B. You mix these solutions together, making solution C. Hint: AgCl is a precipitate.
Calculate the concentrations (in M) of the following ions in solution C.
Na+
To calculate the concentrations of ions in solution C, we need to determine the amount of each ion that is present in the mixture.
In solution A:
- Volume = 3.00 L
- Concentration of NaCl = 3.00 M
In solution B:
- Volume = 2.00 L
- Concentration of AgNO3 = 2.00 M
Since NaCl dissociates into Na+ and Cl- ions in solution, the concentration of Na+ in solution A is also 3.00 M.
Since AgNO3 dissociates into Ag+ and NO3- ions in solution, the concentration of Ag+ in solution B is 2.00 M.
When the two solutions are mixed, a precipitation reaction occurs between the Na+ ions and Ag+ ions, forming AgCl precipitate. The AgCl precipitate is insoluble in water and is removed from the solution.
The amount of Na+ ions in the mixture is equal to the amount of Na+ ions in solution A, which is 3.00 L * 3.00 M = 9.00 mol.
After the precipitation of AgCl, the amount of Ag+ ions in solution C is equal to the amount of Ag+ ions in solution B, which is 2.00 L * 2.00 M = 4.00 mol.
Since the volume of solution C is the sum of the volumes of solutions A and B, the total volume of solution C is 3.00 L + 2.00 L = 5.00 L.
The concentration of Na+ in solution C is therefore 9.00 mol / 5.00 L = 1.80 M.
During the precipitation reaction, all the Ag+ ions form solid AgCl, and no Ag+ ions remain in solution C. Therefore, the concentration of Ag+ in solution C is 0 M.